SOLUTION: Hi, Can i get solution for this problem a^(logb-logc)* b^(logc-loga)* C^(loga-logb) Thanks and regards, AN

Question 478167: Hi,
Can i get solution for this problem
a^(logb-logc)* b^(logc-loga)* C^(loga-logb)
Thanks and regards,
AN
Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
a^(logb-logc)* b^(logc-loga)* C^(loga-logb)
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We can remove the exponents by taking the log of the expression:
log(a^(logb-logc)* b^(logc-loga)* C^(loga-logb)) =
log(a^(logb-logc)) + log(b^(logc-loga)) + log(c^(loga-logb))
= (logb-logc)loga + (logc-loga)logb + (loga-logb)logc
= logb*loga - loga*logc + logc*logb - loga*logb + loga*logc - logc*logb
All terms cancel, so the epxression = 0.
To get back the original expresion, we need to "undo" the log:
a^(logb-logc)*b^(logc-loga)*c^(loga-logb) = 10^0 = 1
Ans: 1
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